Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in PYTHON3):

Solution:
n=int(input())
p=list( float(s) for s in input().split() )
res=[1-p[0], p[0]]+[0]*(n//2 - 1)
for i in range(1,n):
    p1=p[i]
    p0=1-p[i]
    res[len(res)-1]=p0*res[len(res)-1]
    for j in range(len(res)-2, -1, -1):
        res[j+1]+=(res[j]*p1)
        res[j]=(res[j]*p0)        
print( (1-sum(res[:1+n//2])) )